What is the integral of (2x^2-1)/(x^3-5x^2+6x) ?

The integral of (2x^2-1)/(x^3-5x^2+6x) is -1/6 ln|x|-7/2 ln|x-2|+17/3 ln|x-3|+C

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integral of (2x^2-1)/(x^3-5x^2+6x)

\int \frac{2 x ^{2} - 1}{x ^{3} - 5 x ^{2} + 6 x} d x

- \frac{1}{6} ln ∣ x ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + \frac{17}{3} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{2 x ^{2} - 1}{x ^{3} - 5 x ^{2} + 6 x} d x

Expand

\frac{2 x ^{2} - 1}{x ^{3} - 5 x ^{2} + 6 x} : \frac{2 x ^{2}}{x ^{3} - 5 x ^{2} + 6 x} - \frac{1}{x ^{3} - 5 x ^{2} + 6 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x ^{2}}{x ^{3} - 5 x ^{2} + 6 x} d x - \int \frac{1}{x ^{3} - 5 x ^{2} + 6 x} d x

\int \frac{2 x ^{2}}{x ^{3} - 5 x ^{2} + 6 x} d x = 2 (- 2 ln ∣ x - 2 ∣ + 3 ln ∣ x - 3 ∣)

\int \frac{1}{x ^{3} - 5 x ^{2} + 6 x} d x = \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 3 ∣

= 2 (- 2 ln ∣ x - 2 ∣ + 3 ln ∣ x - 3 ∣) - (\frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 3 ∣)

Simplify

2 (- 2 ln ∣ x - 2 ∣ + 3 ln ∣ x - 3 ∣) - (\frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 3 ∣) : - \frac{1}{6} ln ∣ x ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + \frac{17}{3} ln ∣ x - 3 ∣

= - \frac{1}{6} ln ∣ x ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + \frac{17}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + \frac{17}{3} ln ∣ x - 3 ∣ + C

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What is the integral of (2x^2-1)/(x^3-5x^2+6x) ?
The integral of (2x^2-1)/(x^3-5x^2+6x) is -1/6 ln|x|-7/2 ln|x-2|+17/3 ln|x-3|+C